Question

To calculate the volume of the described solid (pyramid).

Short answer

The volume of the pyramid is \(\frac{{\sqrt 3 }}{{12}}{a^2}h\).

Step-by-step solution

Given data

The pyramid with height \(h\) and base \(b\) an equilateral triangle with side \(a\).

Consider the side length at any height is \(\alpha \).

The dimension of the pyramid as shown in Figure 1.

Refer to Figure 1.

The side length is 0 at the height \(h\) and \(a\) at \(h = 0\).

Apply proportion for the cross section.

\(\begin{aligned}\frac{\alpha }{a} &= \frac{{h - y}}{h}\\\alpha &= a\left( {1 - \frac{y}{h}} \right)\end{aligned}\)

The region lies between \(a = 0\) and \(b = h\).

Concept used of pyramid

A pyramid is a 3D figure built with a base of a polygon and triangular faces all connected together. A pyramid connects each vertex of the base to a common tip or apex which gives it the typical shape. Let us learn more about pyramids in this section.

Solve to find the volume

The expression to find the volume of the pyramid as shown below.

\(V = \int_a^b A (y)dy\) …..(1)

Find the area of the equilateral triangle as shown below.

\(\begin{aligned}A(y) &= \frac{{\sqrt 3 }}{4}{\alpha ^2}\\ &= \frac{{\sqrt 3 }}{4}{\left( {a\left( {1 - \frac{y}{h}} \right)} \right)^2}\\ &= \frac{{\sqrt 3 {a^2}}}{4}\left( {1 + \frac{{{y^2}}}{{{h^2}}} - 2\frac{y}{h}} \right)\end{aligned}\)

Substitute 0 for a, h for \(b\), and \(\frac{{\sqrt 3 {a^2}}}{4}\left( {1 + \frac{{{y^2}}}{{{h^2}}} - 2\frac{y}{h}} \right)\) for \(A(y)\) in Equation (1).

\(\begin{aligned}V &= \int_0^h {\frac{{\sqrt 3 {a^2}}}{4}} \left( {1 + \frac{{{y^2}}}{{{h^2}}} - 2\frac{y}{h}} \right)dy\\ &= \frac{{\sqrt 3 {a^2}}}{4}\left( {y + \frac{1}{{{h^2}}}\frac{{{y^3}}}{3} - \frac{2}{h}\frac{{{y^2}}}{2}} \right)_0^h\\ &= \frac{{\sqrt 3 {a^2}}}{4}\left( {h + \frac{1}{{{h^2}}}\frac{{{h^3}}}{3} - \frac{{{h^2}}}{h} - 0} \right)\\ &= \frac{{\sqrt 3 {a^2}}}{4}\left( {h + \frac{h}{3} - h} \right)\end{aligned}\)

Which means, \(V = \frac{{\sqrt 3 }}{{12}}{a^2}h\).

Therefore, the volume of the pyramid is \(\frac{{\sqrt 3 }}{{12}}{a^2}h\).